Apositive point charge q is located at x=a and a negative point charge −q is at x=−a. a positive charge q can be placed anywhere on the y-axis. find an expression for (fnet)x, the x-component of the net force on q. (give your answer in terms of q, q, a, y and constant k.)

Explanation:

The Coulomb's Law gives the force by the charges:

Let us denote the positon of the charge q on the y-axis as 'y'.

The force between 'Q' and'q' is

where Θ is the angle between and x-axis.

whereas

Finally, the x-component of the net force is

i'm only doing this for the points : (

explanation:

i'm only doing this the points

because it says use -2,0,and 2 for x, we can plug those numbers into the equation. y/4=3/4(-2)+2   y/4= -1 1/2 +2 now to get x by itself, we can multiply each side by 4. y=-6+8 which equals y=2 when x=-2. now you can do the same thing with 0 and 2 as x. you get y=8 when x=0 and y=14 when x=2.

x   |   y

-2 |   2

0   |   8

2   |   14